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Tomtit [17]
3 years ago
13

What is an angle that is complementary to DGE? (Please hell I’ll reward)

Mathematics
1 answer:
weqwewe [10]3 years ago
3 0

Answer:

<CGD

Step-by-step explanation:

Since m<CGB = right angle = 90°, therefore:

m<CGE = 90° (linear pair angles)

Complementary angles add up to 90°.

Thus, m<CGD + m<DGE = m<CGE = 90°.

Therefore, the angle that is complementary to <DGE is <CGD

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fast as heck 0.o (i think 170

Step-by-step explanation:

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∆ABC is an isosceles triangle in which angles B and C are congruent.
Brut [27]
Since B and C are congruent, that means that 4x+10=x+34. Using the addition property of equality, I challenge you to solve for x! After you do that, you can note that all the angles in a triangle add up to 180, so 4x+10+x+34+angleA=180. When you find x, you can substitute that in.

If you need help with the addition property of equality for variables, here's an example - 5x+23=4x+3. Subtracting 3 from both sides, we get 5x+20=4x. Next, we can subtract 5x from both sides to get -x=20 and divide both sides by -1 to get x=-20.

Good luck, and feel free to ask with any questions!
4 0
3 years ago
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Find the value of x for the following situation: the line containing (8,7) and (7,-6) is perpendicular to the line containing (2
drek231 [11]

Answer:

x=14

Step-by-step explanation:

7 0
4 years ago
Location is known to affect the number, of a particular item, sold by an automobile dealer. Two different locations, A and B, ar
yKpoI14uk [10]

Answer:

We conclude that the true mean number of sales at location A is fewer than the true mean number of sales at location B.

Step-by-step explanation:

We are given that Location A was observed for 18 days and location B was observed for 13 days.  

On average, location A sold 39 of these items with a sample standard deviation of 8 and location B sold 49 of these items with a sample standard deviation of 4.

<em>Let </em>\mu_1<em> = true mean number of sales at location A.</em>

<em />\mu_2 = <em>true mean number of sales at location B</em>

So, Null Hypothesis, H_0 : \mu_1-\mu_2\geq0  or  \mu_1 \geq \mu_2     {means that the true mean number of sales at location A is greater than or equal to the true mean number of sales at location B}

Alternate Hypothesis, H_A : \mu_1-\mu_2  or  \mu_1< \mu_2    {means that the true mean number of sales at location A is fewer than the true mean number of sales at location B}

The test statistics that would be used here <u>Two-sample t test statistics</u> as we don't know about the population standard deviations;

                        T.S. =  \frac{(\bar X_1-\bar X_2)-(\mu_1-\mu_2)}{s_p\sqrt{\frac{1}{n_1}+\frac{1}{n_2}  } }  ~ t_n__1_-_n__2-2

where, \bar X_1 = sample average of items sold at location A = 39

\bar X_2 = sample average of items sold at location B = 49

s_1 = sample standard deviation of items sold at location A = 8

s_2 = sample standard deviation of items sold at location B = 4

n_1 = sample of days location A was observed = 18

n_2 = sample of days location B was observed = 13

Also,  s_p=\sqrt{\frac{(n_1-1)s_1^{2}+(n_2-1)s_2^{2}  }{n_1+n_2-2} }  = \sqrt{\frac{(18-1)\times 8^{2}+(13-1)\times 4^{2}  }{18+13-2} }  = 6.64

So, <u><em>test statistics</em></u>  =  \frac{(39-49)-(0)}{6.64 \times \sqrt{\frac{1}{18}+\frac{1}{13}  } }  ~ t_2_9  

                               =  -4.14

The value of t test statistics is -4.14.

Now, at 0.01 significance level the t table gives critical value of -2.462 at 29 degree of freedom for left-tailed test.

<em>Since our test statistics is less than the critical values of t as -2.462 > -4.14, so we have sufficient evidence to reject our null hypothesis as it will fall in the rejection region due to which </em><u><em>we reject our null hypothesis</em></u><em>.</em>

Therefore, we conclude that the true mean number of sales at location A is fewer than the true mean number of sales at location B.

3 0
3 years ago
Do you think you can do this
Aleonysh [2.5K]

Mean = Average

High average = +114.2 F

Low average = -36.8 F

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